The output of the circuit is low (zero) for:
(A) \( X = 0, Y = 0 \)
(B) \( X = 0, Y = 1 \)
(C) \( X = 1, Y = 0 \)
(D) \( X = 1, Y = 1 \)
Choose the correct answer from the options given below:
The given circuit consists of two logic gates:
1. The first gate is an AND gate.
2. The second gate is an OR gate. Let's evaluate the output for each pair of \( X \) and \( Y \):
- When \( X = 0 \) and \( Y = 0 \), the output of the AND gate is \( 0 \) because both inputs are zero. The output of the OR gate is also \( 0 \), since the OR gate only outputs 1 when at least one input is 1. Thus, the final output is low (zero). - When \( X = 0 \) and \( Y = 1 \), the output of the AND gate is \( 0 \). The OR gate outputs \( 1 \), but since the AND gate's output is zero, the final output is still low.
- When \( X = 1 \) and \( Y = 0 \), the output of the AND gate is \( 0 \) because the second input is zero. The OR gate outputs \( 1 \), but the final output will still be low.
- When \( X = 1 \) and \( Y = 1 \), the AND gate outputs \( 1 \), and the OR gate also outputs \( 1 \), resulting in a high output.
Thus, the output is low for the following combinations: - (A) \( X = 0, Y = 0 \) - (B) \( X = 0, Y = 1 \) - (C) \( X = 1, Y = 0 \)
Final Answer: (1) (A), (B) and (C) only.
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)